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\[\LARGE y=5^{2x}~ln\sqrt{x}\]
\[ 5^{2x}=25^x \]
\[ \ln\sqrt x=\frac { \ln x}2 \]
Use product rule.
\[\LARGE (2*5^{2x}~ln\sqrt{x})+(5^{2x}*\frac{1}{\sqrt{x}}*\frac{1}{2}x^{-1/2})\] Uh, I think keeping the 5^2x was a bad idea xD
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Yea, I probably screwed that up big time.
Uhm, \[\LARGE \frac{d}{dx}~5^{2x}=2*5^{2x}*ln5\]?
Yes
\[\LARGE (2*5^{2x}*ln5*ln\sqrt{x})+(5^{2x}*\frac{1}{2x})\] Okay there xD
You could have done it the easy way...
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Gotta learn the basic way first I suppose xD
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